Simple harmonic motion and the reference circle

The simplest oscillation

Many oscillations have a displacement that follows a sine or a cosine of time. We give this motion a name. A body is in simple harmonic motion (SHM) when its displacement from the mean position is

Each letter has its own name and meaning.

• $x(t)$ is the displacement at the time $t$.
• $A$ is the amplitude. It is the largest displacement, on either side. The body moves between $x = +A$ and $x = -A$.
• $\omega t + \phi$ is the phase. It tells us where the body is, and which way it is going, at the time $t$.
• $\phi$ is the phase constant. It is the phase at $t = 0$, so it sets where the motion starts.
• $\omega$ is the angular frequency, in rad/s.

The angular frequency

The motion repeats after one period, so the phase grows by $2\pi$ in the time $T$:

We want $\omega$, so $\omega = \dfrac{2\pi}{T} = 2\pi f$. An oscillation of $5\ \text{Hz}$ has $\omega = 2\pi \times 5\ \text{Hz} = 31.4\ \text{rad/s}$.

Look at what the three numbers do to the graph of $x$ against $t$. A larger $A$ makes the curve taller. A larger $\omega$ squeezes the curve, so more oscillations fit into the same time. A change in $\phi$ slides the whole curve along the time axis. It does not change its shape.

A point on a circle

Here is a picture that makes SHM easy to see. Let a point P move round a circle of radius $A$ at a steady angular speed $\omega$. This is uniform circular motion. Let Q be the foot of P on a diameter, so Q is always straight below or above P.

At the time $t$, the radius to P makes the angle $\omega t + \phi$ with the diameter. So Q is at

This is exactly SHM. So the projection of uniform circular motion on a diameter is simple harmonic motion. We call the circle the reference circle. The radius of the circle is the amplitude, and one turn of P is one oscillation of Q.

Think of the second hand of a wall clock. Shine a light from the side, so the shadow of its tip falls on the wall. The shadow goes up and down in SHM, with the period $60\ \text{s}$.

Watch Q in the simulation. Near the centre it moves fast. Near each end it slows down, stops for an instant and turns back. P keeps a steady speed all the time. Only the part of its motion along the diameter changes.

Two worked examples

Example 1. A body moves as $x = 0.08\cos(4\pi t)$, with $x$ in metres. Find the amplitude, the frequency and the period.

The amplitude is $0.08\ \text{m}$. The angular frequency is $\omega = 4\pi\ \text{rad/s}$. We want $f$, so $f = \dfrac{\omega}{2\pi} = \dfrac{4\pi\ \text{rad/s}}{2\pi} = 2\ \text{Hz}$. The period is $T = \dfrac{1}{2\ \text{Hz}} = 0.5\ \text{s}$.

Example 2. A body in SHM starts from $x = 0$ at $t = 0$, moving towards $+x$. What is its phase constant?

At $t = 0$, $x = A\cos\phi = 0$, so $\phi = \pi/2$ or $\phi = -\pi/2$. On the reference circle, P must be at the bottom, moving up and to the right. That is the angle $-\pi/2$. So $\phi = -\pi/2$, and $x = A\cos(\omega t - \pi/2) = A\sin\omega t$.

Practise with new numbers

Try it in Python

The program lists the position of a body in SHM at eight equal steps of one period. Then it finds when the body first reaches half its amplitude.

Common mistakes

• Using degrees in $\cos(\omega t)$. $\omega t$ is in radians. Set your calculator to radians.
• Taking $\omega$ as the frequency. $\omega = 2\pi f$. A motion of $2\ \text{Hz}$ has $\omega = 4\pi\ \text{rad/s}$, not $2\ \text{rad/s}$.
• Taking the amplitude as the whole width. The body moves from $-A$ to $+A$. The whole width of the path is $2A$.
• Forgetting the direction when finding $\phi$. One position at $t = 0$ fits two phase constants. The direction of motion picks one.

Remember these

• SHM: $x = A\cos(\omega t + \phi)$.
• $A$ is the amplitude, $\omega t + \phi$ the phase, $\phi$ the phase constant.
• $\omega = \dfrac{2\pi}{T} = 2\pi f$, in rad/s.
• SHM is the projection of uniform circular motion on a diameter. The radius of the reference circle is the amplitude.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A body moves as $x = 5\cos(2t + \pi/4)$, with $x$ in cm and $t$ in s. Write down its amplitude, angular frequency and phase constant.

2. Find the period of the motion in question 1.

3. A point goes round a circle of radius $20\ \text{cm}$ twice every second. Describe the motion of its foot on a diameter.

4. Two bodies have the same amplitude and period. One has $\phi = 0$, the other $\phi = \pi$. How do their motions compare?